Algebraically closed field: Difference between revisions

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===Polynomials of prime degree have roots===
===Polynomials of prime degree have roots===
If every polynomial over ''F'' of prime degree has a root in ''F'', then every non-constant polynomial has a root in ''F''.<ref>Shipman, J. [http://www.jon-arny.com/httpdocs/Gauss/Shipman%20Intellig%20Mod%20p%20FTA.pdf Improving the Fundamental Theorem of Algebra]  ''The Mathematical Intelligencer'', Volume 29 (2007), Number 4. pp. 9–14</ref> It follows that a field is algebraically closed if and only if every polynomial over ''F'' of prime degree has a root in ''F''.
If every polynomial over ''F'' of prime degree has a root in ''F'', then every non-constant polynomial has a root in ''F''.{{sfn|Shipman|2007}} It follows that a field is algebraically closed if and only if every polynomial over ''F'' of prime degree has a root in ''F''.


===The field has no proper algebraic extension===
===The field has no proper algebraic extension===
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The theory of algebraically closed fields has [[quantifier elimination]].
The theory of algebraically closed fields has [[quantifier elimination]].
There is no algebraically closed [[finite field]]: if there were such a field, with underlying set <math>\{0, 1, k_2, ..., k_n\}</math> for some <math>n\in\N</math>, then the polynomial <math>x(x-1)(x-k_2)...(x-k_n) +1</math> would never vanish for any value of <math>x</math>.


== See also ==
== See also ==
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{{DEFAULTSORT:Algebraically Closed Field}}
{{DEFAULTSORT:Algebraically Closed Field}}
[[Category:Field (mathematics)]]
[[Category:Field theory]]